題目：

Given a string S of '(' and ')' parentheses, we add the minimum number of parentheses ( '(' or ’)', and in any positions ) so that the resulting parentheses string is valid.
Formally, a parentheses string is valid if and only if:
　It is the empty string, or
　It can be written as AB (A

concatenated with B), where A and B are valid strings, or
　It can be written as (A),where A is a valid string.
Given a parentheses string, return the minimum number of parentheses we must add to make the resulting string valid.
Example 1:

Input: "())" 
Output: 1 

Example 2:

Input: "(((" 
Output: 3 
 

Example 3:

Input: "()" 
Output: 0 

Example 4:

Input: "()))((" 
Output: 4

解釋：
需要補充幾個括號使得括號組成的string變為valid，看到括號的題目就想用stack做怎麼辦。用傳統的括號匹配問題，需要用一個count變數記錄沒有匹配的)的個數，最後返回stack的長度+count即可，其中stack的長度表示沒有匹配的(的個數。
python程式碼：

class Solution(object):
    def minAddToMakeValid(self, S):
        """
        :type S: str
        :rtype: int
        """
 

        stack=[]
        count=0
        for letter in  S:
            if letter=='(':
                stack.append(letter)
            else:
                if stack:
                    stack.pop(-1)
                else:
                    count+=1
        return len(stack)+count
        

實際上不用棧，也用另一個計數器記錄(的個數即可，其實也是棧的思想。
python程式碼：

class Solution(object):
    def minAddToMakeValid(self, S):
        """
        :type S: str
        :rtype: int
        """
        left,right=0,0
        for letter in S :
            if letter=='(':
                left+=1
            else:
                if left>0:
                    left-=1
                else:
                    right+=1
        return left+right 

c++程式碼：

 class Solution {
public:
    int minAddToMakeValid(string S) {
        int left=0,right=0;
        for(auto letter:S)
        {
            if (letter=='(')
                left++;
            else
            {
                if(left)
                    left--;
                else
                    right++;
            }
            
        }
        return left+right;
    }
};

總結：